Question: As $t$ takes on all real values, the set of points $(x,y)$ defined by
\begin{align*}
x &= t^2 - 2, \\
y &= t^3 - 9t + 5
\end{align*}forms a curve that crosses itself.  Compute the ordered pair $(x,y)$ where this crossing occurs.
Suppose the curve intersects itself when $t = a$ and $t = b,$ so $a^2 - 2 = b^2 - 2$ and $a^3 - 9a + 5 = b^3 - 9b + 5.$  Then $a^2 = b^2,$ so $a = \pm b.$  We assume that $a \neq b,$ so $a = -b,$ or $b = -a.$  Then
\[a^3 - 9a + 5 = (-a)^3 - 9(-a) + 5 = -a^3 + 9a + 5,\]or $2a^3 - 18a = 0.$  This factors as $2a (a - 3)(a + 3) = 0.$

If $a = 0,$ then $b = 0,$ so we reject this solution.  Otherwise, $a = \pm 3.$  For either value, $(x,y) = \boxed{(7,5)}.$